Question: Let $m=x^2-5$. Which equation is equivalent to $(x^2-5)^2-3x^2+15=-2$ in terms of $m$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $m^2-3m+2=0$ (Choice B) B $m^2+3m+2=0$ (Choice C) C $m^2+3m+17=0$ (Choice D) D $m^2-3m+17=0$
Solution: We are asked to rewrite the equation in terms of $m$, where ${m}={x^2-5}$. In order to do this, we need to find all of the places where the expression ${x^2-5}$ shows up in the equation, and then substitute ${m}$ wherever we see them! For instance, note that $-3x^2+15=-3({x^2-5})$. This means that we can rewrite the equation as: $(x^2-5)^2-3x^2+15=-2$ $({x^2-5})^2-3({x^2-5})=-2$ [What if I don't see this factorization?] Now we can substitute ${m}={x^2-5}$ : $({m})^2-3({m})=-2$ Finally, let's manipulate this expression so that it shares the same form as the answer choices: ${m}^2-3{m}+2=0$ In conclusion, $m^2-3m+2=0$ is equivalent to the given equation when $m=x^2-5$.